The equilibrium constant of a reaction at 298 K is 1.0 x 10^-3. Using 2.303RT = 5.71 kJ mol^-1, Delta G deg is

JEE Main levelSingle correct optionDifficulty 2 of 3td099

The equilibrium constant of a reaction at $298 K$ is $1.0 \times 1 0^{- 3}$ . Using $2.303 RT = 5.71 kJ mo l^{- 1}$ , $Δ G^{\circ}$ is

Δ G^{\circ} = - 2.303 RT lo g_{10} K = - 5.71 \times (- 3) = + 17.1 kJ mo l^{- 1}

. A small

K

corresponds to a positive

Δ G^{\circ}